Thursday, February 7, 2008

Lecture 8: Hardness of Max-Ek-Independent-Set

Any questions about today's lecture? I will repeat here the hardness reduction:

Given: Max-Label-Cover(K,L) instance $G = (U,V,E)$ with projection constraints $\pi_{v \to u}$.

Reduction to a weighted k-uniform hypergraph $H$:
. The vertex set is $V \times \{0,1\}^L$ (there are $|V| 2^{|L|}$ vertices). We call each $\{v\} \times \{0,1\}^L$ the "block" of vertices corresponding to $v \in V$. We name the strings/sets in this block $A^{(v)}$.
. The weight on each vertex $A^{(v)}$ is its $p$-biased weight, $p = 1 - 2/k - \delta$. The total weight in each block is $1$ and the overall total weight is $v$.
. For each pair of vertices $v, v' \in V$ sharing a common $u$ (i.e., with $(u,v), (u,v') \in E$), we will put some hyperedges on vertices in the union of the $v$ and $v'$ blocks.
. Specifically, we put a hyperedge on the $k$-set $\{A_1^{(v)}, ..., A_{k/2}^{(v)}, B_{1}^{(v')}, ... B_{k/2}^{(v')}\}$ iff the following condition holds:

$\pi_{v\to u}(\cap A_{i}^{(v)})$ is disjoint from $\pi_{v' \to u}(\cap B_{i}^{(v')})$.

(Supertechnical note. The A's should all be distinct, and the B's should all be distinct. Also, we allow $v = v'$, in which case the A's and B's should be mutually distinct.)

This construction is polynomial time (since $k$, $|K|$, $|L|$ are all constants; supersupertechnical note: we assume $\delta$ is rational).

You should again try to convince yourself of the "completeness" part of the proof: Opt(G) = 1 implies Opt(H) $\geq p = 1 - 2/k - \delta$.

No comments: